Đáp án:
$\begin{array}{l}
+ \cos 2x + \sin 3x = 0\\
\Leftrightarrow \cos 2x = - \sin 3x\\
\Leftrightarrow \cos 2x = \sin \left( {3x + \pi } \right)\\
\Leftrightarrow \cos 2x = \cos \left( {\dfrac{\pi }{2} - 3x - \pi } \right)\\
\Leftrightarrow \left[ \begin{array}{l}
2x = - \dfrac{\pi }{2} - 3x + k2\pi \\
2x = \pi + \dfrac{\pi }{2} + 3x + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{\pi }{{10}} + \dfrac{{k2\pi }}{5}\\
x = - \dfrac{{3\pi }}{2} - k2\pi
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = - \dfrac{\pi }{{10}} + \dfrac{{k2\pi }}{5}\\
x = - \dfrac{{3\pi }}{2} - k2\pi
\end{array} \right.\\
+ )\tan x - \cot 3x = 0\\
\Leftrightarrow \tan x = \cot 3x\\
\Leftrightarrow \tan x = \tan \left( {\dfrac{\pi }{2} - 3x} \right)\\
\Leftrightarrow x = \dfrac{\pi }{2} - 3x + k\pi \\
\Leftrightarrow x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}\\
Vậy\,x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{4}
\end{array}$
